设x₁= sinθ,x₂=cosθ.
由韦达定理,x₁+ x₂= k,x₁·x₂=k+1.
∴ x₁²+ x₂²=1=( x₁+ x₂)²-2 x₁·x₂=k²-2(k+1),解得k=-1或3.
又△=k²-4(k+1)≥0.∴k=-1.
此时方程化为x²+x=0,解得x=-1或0,故①sinθ=-1,cosθ=0 ②sinθ=0,cosθ=-1.
由①得θ=3π/2,由②得θ=π.